Question: $g(t) = -2t^{3}+t^{2}+t+1+f(t)$ $f(x) = x^{2}-3x$ $ g(f(0)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = 0^{2}+(-3)(0)$ $f(0) = 0$ Now we know that $f(0) = 0$ . Let's solve for $g(f(0))$ , which is $g(0)$ $g(0) = -2(0^{3})+0^{2}+1+f(0)$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = 0^{2}+(-3)(0)$ $f(0) = 0$ That means $g(0) = -2(0^{3})+0^{2}+1$ $g(0) = 1$